# Is $\omega_1$ metrizable?

Following Urysohn's metrization theorem, I would like to prove or disprove that $\omega_1$ is metrizable. I know it is hausdorff, but I'm not sure whether or not it is second countable, and I'm at loss on how I can prove or disprove it is second countable. And if I'll disprove it, will it tell me that $\omega_1$ is not metrizable, or is Urysohn's theorem only an 'if' and not an 'iff'?

Or perhaps there is a more comfortable-to-use in here theorem to prove/disprove $\omega_1$ being metrizable?

(1) $$\omega_1$$ is not only Hausdorff, it is normal, as is every linearly ordered set in its order topology.

(2) $$\omega_1$$ is not Lindelöf, and thus not second countable, because the intervals $$[0,\alpha)$$ for $$\alpha\lt\omega_1$$ are an open cover with no countable subcover.

(3) $$\omega_1$$ is not separable. Proof: Let S be any countable subset of $$\omega_1$$. Then $$\sup S\lt\omega_1$$. Let $$\alpha=\sup S$$. Then $$S$$ is a subset of the closed set $$[0,\alpha]$$, so the closure of $$S$$ is a subset of $$[0,\alpha]$$ which is a proper subset of $$\omega_1$$, so $$S$$ is not dense in $$\omega_1$$.

(4) $$\omega_1$$ is sequentially compact. Proof: Every sequence in $$\omega_1$$ has a monotonic subsequence, which converges.

(5) If a metric space is not separable, then it is not sequentially compact. Proof: Let $$X$$ be a metric space. If $$X$$ is not separable, then for some $$\varepsilon\gt0$$ there is an uncountable set of points such that the distance between any two is at least $$\varepsilon$$. An infinite sequence of those points can not converge, and it can not have a convergent subsequence. Therefore $$X$$ is not sequentially compact.

From (3), (4), and (5) it follows that $$\omega_1$$ is not metrizable. Namely, if $$\omega_1$$ were metrizable, then, since by (3) it is not separable, it would follow by (5) that it is not sequentially compact, contradicting (4).

• Just a couple of comments. (1) An alternative way of showing that $\omega_1$ is not second-countable is to note that $\{ \alpha+1 \}$ is open in $\omega_1$ for all $\alpha < \omega_1$, and so any base for $\omega_1$ must contain all of these uncountably many singletons. (2) In metrizable spaces the notions of compactness, countable compactness, and sequential compactness are equivalent. (So $\omega_1$ is not metrizable since it is sequentially, and countably, compact, but not compact, or even Lindelöf.) Oct 17, 2014 at 14:31
• @ArthurFischer Thanks for your comments. Also $\omega_1$ is not metrizable because it's not paracompact or even meta-Lindelöf. A subspace $X$ of $\omega_1$ is metrizable if and only if $X$ is a nonstationary subset of $\omega_1$.
– bof
Oct 17, 2014 at 21:32